mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-11-30 16:31:08 +00:00
51 lines
1.2 KiB
Python
51 lines
1.2 KiB
Python
"""
|
|
* Binary Exponentiation with Multiplication
|
|
* This is a method to find a*b in a time complexity of O(log b)
|
|
* This is one of the most commonly used methods of finding result of multiplication.
|
|
* Also useful in cases where solution to (a*b)%c is required,
|
|
* where a,b,c can be numbers over the computers calculation limits.
|
|
* Done using iteration, can also be done using recursion
|
|
|
|
* @author chinmoy159
|
|
* @version 1.0 dated 10/08/2017
|
|
"""
|
|
|
|
|
|
def b_expo(a, b):
|
|
res = 0
|
|
while b > 0:
|
|
if b&1:
|
|
res += a
|
|
|
|
a += a
|
|
b >>= 1
|
|
|
|
return res
|
|
|
|
|
|
def b_expo_mod(a, b, c):
|
|
res = 0
|
|
while b > 0:
|
|
if b&1:
|
|
res = ((res%c) + (a%c)) % c
|
|
|
|
a += a
|
|
b >>= 1
|
|
|
|
return res
|
|
|
|
|
|
"""
|
|
* Wondering how this method works !
|
|
* It's pretty simple.
|
|
* Let's say you need to calculate a ^ b
|
|
* RULE 1 : a * b = (a+a) * (b/2) ---- example : 4 * 4 = (4+4) * (4/2) = 8 * 2
|
|
* RULE 2 : IF b is ODD, then ---- a * b = a + (a * (b - 1)) :: where (b - 1) is even.
|
|
* Once b is even, repeat the process to get a * b
|
|
* Repeat the process till b = 1 OR b = 0, because a*1 = a AND a*0 = 0
|
|
*
|
|
* As far as the modulo is concerned,
|
|
* the fact : (a+b) % c = ((a%c) + (b%c)) % c
|
|
* Now apply RULE 1 OR 2, whichever is required.
|
|
"""
|